fasteners lecture 11-20-03

8/3/2019 Fasteners Lecture 11-20-03

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ME 351

Power Screws, Fasteners,

and Connections


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Threads and Connections

Today we will start off discussing the

mechanics of screw threads. Next, powerscrews & threaded fasteners will be

examined. Since threaded fasteners are

often used to make connections, we will end

with that topic.


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The Inclined Plane

Truly one of the worlds great inventions!

By inspection, a steeper angle gains youelevation more quickly, but the appliedforce must increase.

QfN

N

W


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Helically-Inclined Planes

Differential element of one thread transferring force tothe mating thread. The helix or lead angle = the slope

of the ramp, and is a critical design parameter. is the

thread angle, and is another important parameter.


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, , andf On a screw thread, the helix angle

controls the distance traveled per revolutionand the force exerted.

, the thread angle, effects the friction forceresisting motion. Sometimes friction isdesirable (e.g., so that threads wontloosen), and sometimes it is not.

fis the coefficient of friction, and plays animportant role in all threads.


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and

, the helix angle, is given by

tan = L/(dm) (eq. 15.2)where,

L = the lead or pitch (threads per unit length)

dm = the mean dia. of the thread contactsurface.

, the thread angle, is determined by thedesign of the threads.


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Thread Friction Examples

Acme Threads

Bolt Threads

Pipe Threads


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Power

ScrewsForce F acts on

moment arm a to

produce a torque T.

Table 15.3 in the text

shows standard sizes

of power screw

threads.

In this drawing, only

the nut rotates.


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Power Screw Thread Types

Acme: in

wide use, but

less efficient.

Square: most

efficient, but

hard to make.

Modified Square:

compromise.


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Power Screws

Equations 15.6 through 15.13 in the text are

the governing equations for torque and

efficiency, given the geometry of thethreads.

However, as in the case of many previous

problems, often you are presented with thatinformation and must solve for other

variables.


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Power Screw Efficiency

Plot of equation

15.13; note thewide range as a

function of both

fand .


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Problem

15.62 square thread

power screw lifts W of

50 kips at 2 fpm.

Find rpm n, and the

HP required if the

efficiency is 85%, and

f= 0.15.

Neglect the collar

friction.


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Threaded Fasteners

Nomenclature


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Threaded Fasteners

Thread Forms

Note that the crests & roots may be either flat or rounded


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Threaded Fasteners: UNS & ISO

UNS = Unified National Standard. Threads arespecified by the bolt or screw diameter (alsocalled the major diameter)in inches, and the

number of threads per inch. ISO = International Standards Organization.

Threads are specified by the major diameter in

mm, and the pitch, or, number of mm perthread.

Generally UNS and ISO threads are NOTinterchangeable. (3mm is close to 1/8.)


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Threaded FastenersUNS

The specification is written in the format

Dia threads/inUNC or UNFclass andinternal or externalRH or LH.

UNC = Unified National Coarse

UNF = Unified National Fine

Class ranges from 1 (cheap & inaccurate) to 3

(expensive & precise). Class 2 is common.

A = external, B = internal


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Threaded FastenersUNS

RH = right hand threads, LH = left hand

Example thus would be:

13 UNC2ARH

Notes:

1. UNF and UNC are redundant information.

2. For diameters less than , a numeric size isspecified instead of the diameter. (00012?)

3. Summarized on page 602 in text, Table 15.1


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Threaded FastenersISO

Metric designations are a little simpler.Preceded by an M, then the diameter in

mm, then the pitch (mm per thread, notthreads per mm). There are also coarse andfine threads in the ISO system.

Examples: M10 x 1.5

M10 x 1.25


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ME 351

Coarse Versus Fine Threads

Coarse threads are fine for normalapplications. They are easier to assemble, a

little more forgiving of dings, possibly cheaperto make, and for a given size of bolt, they exertless force than do fine threads; good for softermaterials bolted together.

Fine threads develop greater force per appliedtorque, and are more effective at resistingvibration-induced loosening.


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Bolts, Screws, and Studs

The same fastener could be a bolt or a screw, depending on

if a nut is used. Studs are threaded at both ends.


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Bolt Grades

Bolts (and nuts) are made from a variety of

materials. The SAE Grade is an indication

of the strength of the material, based on theproof stress, Sp (slightly less than the yield

stress). Sp ranges from 33 ksi for a grade1

bolt, up to 120 ksi for a grade 8 bolt. Theproof loadof a bolt is the load at which

permanent deformation commences.


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SAE Bolt Head Markings

SAE 2 SAE 5 SAE 7 SAE 8

http://raskcycle.com/techtip/webdoc14.html

Hexagonal bolt heads are stamped with radial lines to

indicate the grade. The grade = the number of lines + 2.


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Thread Manufacture

Threads are generally produced by either

rolling (forming with a specialized die) or

by cutting, as on a lathe. Rolled threads arestronger and have better fatigue properties

due to the cold work put into the material.

Power screw threads may be ground toachieve a very smooth surface to reducef.

Threads may also be cast into a part.


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ME 351

Stresses in Threaded Fasteners

Due to imperfect thread spacing, most of the

load between a bolt and a nut is taken by the

first pair of threads. This is partiallyrelieved by bending and localized yielding,

however most thread failures occur in that

region. The stress concentration rangesfrom 2 to 4.


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Major-Diameter Stresses

Axial stress is given by the familiar

= P/A

For A, use either the root diameter for powerscrews, or tabulated values for fasteners.

Torsional stress is given by the familiar = T/J = 16T/d3

See p. 615 for interpretation of T and d. T isthe applied torque for power screws, or thewrench torque, for fasteners.


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Bearing Stress

Bearing stress, the compressive stressbetween the surfaces of the threads, is givenby b = P/(dmhne) (eq. 15.17)

P = load,

dm = pitch or mean screw thread diameter,

h = depth of thread, and

ne = number of threads in engagement.

b is usually not a limiting design factor.


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Nomenclature for Thread

Stress Analysis


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Direct Shear Stress on Threads

Then we have,

= 3P/(2 dbne), where,

d = root dia. for the screw or major dia. for thenut,

b = the thread thickness at the root, and

ne = the number of threads in engagement.

Note that can be a limiting factor.


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Bolt Tightening & Preload

Bolted joints commonly hold parts together

in opposition to both normal and shear

forces.In certain applications it is desirable to

tighten a bolted joint to a specified preload

Fi, which is some fraction of the bolts proofload, Fp.


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Bolt Tightening & Preload

An engineer would specify a preload in the

case of fatigue applications, in order to

minimize the relative magnitude of thealternating load Pa compared to the average

load Pmean. (Recall definitions from ch. 8.)

Preloading is also important in sealingapplications, as in a gasketed joint. Both

reasons are important for auto cylinder heads.


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Preload Values

The optimum preload is often given by eq. 15.20:

Fi = 0.75 Fp for connections to be reused, or

Fi = 0.90 Fp for permanent connections.

The proof load Fp is found from eq. 15.14 as,

Fp

= SpA

t, where the proof stress S

pis an SAE

specification (see Table 15.4 or 15.5), and tension

area At is found in Table 15.1 or 15.2.


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Tightening Torque

To develop the specified preload, the

tightening torque is given by eq. 15.21:

T = KdFi, whereT = the tightening torque,

d = the nominal bolt diameter (e.g., ),

Fi = the desired preload, andK = a torque coefficient


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Tightening Torque

Equation 15.21 is approximate, and applies

for standard threads.

For dry, unlubricated, or average threads,

K = 0.2. For lubricated threads, K = 0.15.

Rewrite eq. 15.21 as,

Fi = T/(Kd) to see that, for a given torque, Fiincreases with lubricated threads.


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Relaxation and Exactness

Most joints lose on the order of 5% of the

original preload over time, due to relaxation

effects (usually over the course of 100s or1000s of hours).

By now it should be clear that threaded

fasteners are extremely complex. Oftenextensive testing is done for critical

applications.


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Tension Joints

Bolted joints are frequently used to clamp

together parts that themselves carry

additional loads: these additional loadsincrease the bolt tension. The engineer

often must determine acceptable loads for

such joints.We consider both the joints and the parts as

springs, with spring constants kb and kp.


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Tension Joints

After assembly with preload Fi, applied load P

will change the force in the bolt and the parts.


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Tension Joints

P = Fb + Fp, where

Fb = the increased tension in the bolt, and

Fp = the decreased compression force in the

parts. The deformations are given by

b = Fb/kb, and p = Fp/kp

Then compatibility requires that

Fb/kb = Fp/kp


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Joint Constant C

The joint stiffness factor, or joint constant, is

defined in eq. 15.22 as C = kb/(kb + kp).

Then the preceding equations yieldFb = CP and Fp = (1C)P

kb is usually small compared to kp, and so C is

a small fraction.


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Forces in Bolted Joints

When a load P is applied to a bolted joint, the

tensile force Fb in the bolt increases, and the

compressive force Fp in the parts decreases.As long as Fp > 0, the forces are:

Fb = CP + Fi (eq. 15.23)

and,

Fp = (1C)PFi (eq. 15.24)


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Determination of C

Review from Chapter 4 about 100 years ago:

Deflection is given by = PL/AE, and the

spring rate k is given by k = P/.Combining these we obtain

kb = AbEb/L (eq. 15.31),

and,

kp = ApEp/L (eq. 15.32)


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Determination of kb

In determining kb, the

threaded and the unthreaded

parts of the bolt are

considered as separate

springs in series. Equation

15.33 gives:

1/kb = Lt/AtEb + Ls/AbEb


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Determination of kp

kp is more complex:

the stress distribution

in the parts is clearly

non-uniform, anddepends on factors

like washers, etc.

It is approximated bythe double-cone

illustrated.


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Determination of kp

Estimate of kp for standard hex-head bolts and

washers is given by eq. 15.34:

kp = (.5 Epd)/{2 ln [5(.58L+.5d)/(.58L+2.5d)]}

d = bolt diameter and

L = grip (thickness of bolted assembly).

Alternatively, just use kp = 3kb !


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Example: Problem 15.11

Bolt diameter is 15 mm.

Grip length L = 50 mm.

Tightening torque foraverage threads is

T = 72 N-m by eq. 15.21

Find maximum P thatwill not loosen the initial

compression in the part


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Example: Problem 15.16

Bolt is M20 x 2.5 coarse thread.

Sy is 630 MPa.

Ep = Eb

L = 60 mm, and P = 40 kN

Determine:

Total force on bolt if joint is

reusable, and, the tightening torque

if the threads are lubricated.


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Typo!

On page 625, the equation for Pa is

incorrect. It should read,

Pa = (PmaxPmin)

(This is in section 15.12, Tension Joints

Under Dynamic Loading.)


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Some Rules of Thumb for

Threaded Fasteners Threaded depth: for a bolt diameter d, the

length of full thread engagement should be

1.0din steel, 1.5din cast iron, and 2.0dinaluminum.

In gasketed joints, bolts are arrayed in a bolt

circle or other pattern. The bolt-to-boltspacing should not exceed about 6dto

maintain uniform pressure.


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Rivets

Rivets often find application in larger structures such

as bridges and towers. They are also used

extensively in aircraft construction. A rivet starts offas a cylinder with one head (usually rounded). The

protruding cylinder is deformed to create a second

head, which locks the joint in compression.


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Joints Primarily in Shear

Both bolts and rivets are used in

connections that primarily experience shear

loading (separate from the case of axial ornormal loading which we just examined).

Such connections may experience any of

several failure modes, and the engineermust analyze for each mode.


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Shear Joint Failure Modes

Shearing Failure of Fastener: = 4P/d2

d = diameter of fastener (from Table 15.7)


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Tensile Failure of Plate: t= P/(wd

e)t, where

de = effective hole dia., w = width, and t = thickness

of thinnest plate (from Table 15.7).


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Effective Hole Diameter

In analyzing potential tensile failure of the

plate, the effective hole diameter is used

rather than the diameter of the fastener.de= the fastener diameter + 1/16 for drilled

holes, or,

de= the fastener diameter + 1/8 forpunched holes (this is usually used).


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Bearing Failure of Plate or Fastener: b = P/dt, where d =diameter of fastener and t = thickness of the thinnest

plate. (from Table 15.7)


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Shearing Failure of Plate: t = P/2at, where t =

thickness of thinnest plate and a = closest distance

from fastener to edge. (from Table 15.7)

a >= 1.5d


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Joint Efficiency

The efficiency of a joint is defined as:

e = Pall/Pt, (eq. 15.41)

where

Pall is the smallest of the allowable loads in

the preceding failure mode examples, and

Pt is the static tensile strength of the plate with

no holes. e is always less than 100%.


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Shear Example, Problem 15.22

Plate thickness is

3/8. Rivets are

diameter, holes

drilled 2 apart.Sall,tension = 22ksi;

Sall,bearing = 48 ksi, and

all = 15 ksi.

Find the joint

efficiency.


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Welded Joints

Welded joints are produced by localized

melting of the parts to be joined, in the

region of the joint. Often a filler metal (orplastic, in the case of plastics) is added,

creating a chemical bond in the parts that

may be stronger than the base material.There are many, many welding processesan

entire engineering major.


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Strength of Butt Welds

The height h does not include the crowned region;

generally it is just the plate thickness.


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Strength of Fillet Welds

Specified size is based on h, but stress is calculated

with t, the region of minimum cross sectional area.


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Factor of Safety for

Welds in ShearJust as many riveted or bolted joints are in

shear, so too are many welded joints. The

factor of safety for a welded joint is givenby:

n = Sys/ = 0.5Sy/ (eq. 15.44)


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Eccentric Loading of Welded

Joints (P.E. Question!)Determination of the exact stress

distribution is very complicated.

With some simplifying

assumptions, the following

procedure gives reasonably

accurate results.

Direct shear stress is given by

d = P/A, where A = the throat

area of all the welds.


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Eccentric Loading

of Welded Jointsd is taken to be uniformly distributed over thelength of all the welds.

Due to the eccentricity e, a torque T is developedabout the centroid C of the weld group: T = Pe.The torque causes an additional shear stress inthe welds:

t = Tr/J (eq. 15.46)

J = polar moment of inertia of the weld groupabout C, based on the throat area. (Continued:)


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Eccentric Loading

of Welded Jointst = Tr/J (eq. 15.46)

In this equation, r is the distance from C to

the point in the weld of interest. t is notuniform across the weld group, and one

point will experience the greatest stress

resultant: = (t

2 + d2) (eq. 15.47)


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Location of the

Centroid (Review)C is located at coordinates x-bar

and y-bar, where

x-bar = (Aixi)/Ai, and

y-bar = (Aiyi)/Ai, where i

denotes a given weld segment,

and the coordinate origin is

conveniently chosen. A key is

that the weld throat t is assumed

to be very small, sometimes 0.


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Moment of Inertia of a Weld and

the Parallel Axis Theorem.Use the familiar bh3/12,

substituting t and L for b and h as

appropriate. However, assume t3

= 0 to simplify.

Remember the parallel axis

theorem, Ix = Ix + Ay12, to find the

moment of inertia about thecentroid of the weld group. (So

even if Ix = 0, you still have A.)


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Polar Moment of Inertia

The polar moment of inertia is the sum of Ixand Iy for each weld about the centroid ofthe weld group. Knowing J, apply

t = Tr/J (eq. 15.46)to find t at a given point, and then use

= (t2 + d

2) (eq. 15.47)

to find the max , which is used to find therequired weld size.

fasteners lecture 11-20-03

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